We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
201
1
avatar

Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?

 Dec 30, 2018
 #1
avatar+5088 
+3

\(\text{Let's say we have an arrangement of lamps such that}\\ \text{the leftmost lamp is blue}\\ \text{and the rightmost lamp is red}\\ \text{Now we select 3 lamps at random and turn them on}\\ \text{There are a total of }\dbinom{6}{3}=20 \text{ ways they can be chosen}\\ \text{If we insist on the left lamp being off and the right one being on}\\ \text{we have }\dbinom{4}{2}=6 \text{ ways of selecting the remaining lamps that will go on}\\ \text{So given an arrangement as described above}\\ P[\text{choosing random 3 lamps to turn on results in left off and right on}]=\dfrac{6}{20}=\dfrac{3}{10}\)

 

\(\text{Now we need to determine the probability of}\\ \text{having the left lamp blue and right lamp red}\\ \text{We fix those two and for the remaining slots we have}\\ \dbinom{4}{2} = 6 \text{ way to select slots for say the red lamps}\\ \text{and having done that the blue lamps are determined}\\ \text{so there are 6 total arrangements satisfying the requirement}\\ \text{out of a possible }\dbinom{6}{3}=20 \text{ ways to arrange the lamps, so}\\ P[\text{leftmost lamp is blue, rightmost lamp is red}]=\dfrac{6}{20}=\dfrac{3}{10}\)

 

\(\text{The probability we're after is just the product of these two}\\ P[\text{left blue lamp off and right red lamp on}]=\left(\dfrac{3}{10}\right)^2=\dfrac{9}{100}\)

.
 Dec 30, 2018
edited by Rom  Dec 30, 2018

17 Online Users

avatar
avatar